Slide 4-1. (a) (NH 4 ) 2 SO 4 (s) 5.0 mol (NH 4 ) 2 SO 4 x. (c) Cu(NO 3 ) 2 (s) 7.42 x formula units x Cu(NO 3 ) 2.

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1 Three Major Classes of Chemical Reactions Chapter 4 Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions 4.4 Acid-Base Reactions 4.5 xidation-reduction (Redox) Reactions 4.6 Elements in Redox Reactions Slide 4-1 Slide 4-2 Electron distribution in molecules of H 2 and H 2. As a result, water is a polar solvent and is able to solvate polar molecules and ionic species Determining Moles of Ions in Aqueous Ionic Solutions How many moles of each ion are in the following solutions? (a) 5.0 mol of ammonium sulfate dissolved in water (b) 78.5 g of cesium bromide dissolved in water (c) 7.42 x formula units of copper(ii) nitrate dissolved in water (d) 35 ml of 0.84 M zinc chloride (a) (NH 4 ) 2 S 4 (s) 2NH 4 + (aq) + S 4 2- (aq) 5.0 mol (NH 4 ) 2 S 4 x 2 mol NH mol (NH 4 ) 2 S 4 = 10. mol NH 4 + and 5.0 mol S 4 2- (c) Cu(N 3 ) 2 (s) Cu 2+ (aq) + 2N 3 - (aq) 7.42 x formula units x Cu(N 3 ) 2 mol Cu(N 3 ) x formula units = mol Cu(N 3 ) 2 = mol N - 3 Slide 4-3 and mol Cu 2+ Slide 4-4 Writing Equations for Aqueous Ionic Reactions The molecular equation shows all of the reactants and products as intact, undissociated compounds. The total ionic equation Writing Equations for Aqueous Ionic Reactions The molecular equation shows all reactants and products as if they were intact, undissociated compounds. This gives the least information about the species in solution. 2AgN 3 (aq) + Na 2 Cr 4 (aq) Ag 2 Cr 4 (s) + 2NaN 3 (aq) shows all of the soluble ionic substances dissociated into ions. The net ionic equation omits the spectator ions and shows the actual chemical change taking place. When solutions of silver nitrate and sodium chromate mix, a brick-red precipitate of silver chromate forms. Slide 4-5 Slide 4-6

2 The total ionic equation shows all soluble ionic substances dissociated into ions. The net ionic equation eliminates the spectator ions and shows only the actual chemical change. This gives the most accurate information about species in solution. 2Ag+ (aq) + 2N3- (aq) + 2Na+ (aq) + Cr42- (aq) Ag2Cr4 (s) + 2Na+ (aq) + 2N3- (aq) 2Ag+ (aq) + Cr42- (aq) Ag2Cr4 (s) Spectator ions are ions that are not involved in the actual chemical change. Spectator ions appear unchanged on both sides of the total ionic equation. 2Ag+ (aq) + 2N3- (aq) + 2Na+ (aq) + Cr42- (aq) Ag2Cr4 (s) + 2Na+ (aq) + 2N3- (aq) Slide 4-7 Slide 4-8 An aqueous ionic reaction and the three types of equations. Figure 4.6 The reaction of Pb(N3)2 and NaI. NaI(aq) + Pb(N3)2(aq) PbI2(s) + NaN3(aq) 2NaI(aq) + Pb(N3)2(aq) PbI2(s) + 2NaN3(aq) 2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2N3-(aq) PbI2(s) + 2Na+(aq) + 2N3-(aq) 2NaI(aq) + Pb(N3)2(aq) PbI2(s) + 2NaN3(aq) Double-displacement reaction (metathesis) Slide 4-9 Predicting Whether a Precipitate Will Form Slide 4-10 Solubility Rules 1. Note the ions present in the reactants. 2. Consider the possible cation-anion combinations (Double Displacement). 3. Decide whether any of the ion combinations is insoluble. To do this we need to know the solubility rules. Slide 4-11 Slide 4-12

3 Predicting Precipitation Reactions Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) Potassium fluoride(aq) + strontium nitrate(aq) (b) Ammonium perchlorate(aq) + sodium bromide(aq) Acids and Bases: Definitions-1 (a) KF(aq) + Sr(N 3 ) 2 (aq) 2K + (aq) + 2F - (aq) + Sr 2+ (aq) + 2N 3 - (aq) 2KN 3 (aq) + SrF 2 (s) 2K + (aq) + 2N 3 - (aq) + SrF 2 (s) (b) NH 4 Cl 4 (aq) + NaBr (aq) NH 4 Br (aq) + NaCl 4 (aq) All reactants and products are soluble so no reaction occurs. All definitions are operationally equivalent; however the Lewis definition is by far the most general 14 Slide 4-13 Slide 4-14 Acids and Bases: Definitions-2 The H + ion as a solvated hydronium ion. H +, as a discrete species, does NT exist in water; rather, it combines with water to give hydronium ion: It is the magnitude of the hydronium ion concentration ([H 3 + ]; in mol/l) that determines how acidic a solution is! Since H 3 + is generated from H +, and since H + comes directly from the acid under consideration, measurement of the [H 3 + ] will have a direct correlation to the strength of the acid! H + interacts strongly with H 2, forming H 3 + in aqueous solution (NT H + ). Slide 4-15 Slide 4-16 Acids and Bases: Definitions-3 PRTN: an archaic term still used to describe a hydrogen ion ( H + ); in acid/base chemistry it is still used to denote an acidic hydrogen and should NT be confused with the more traditional definition of a proton (a positively charged nuclear particle). PRTN TRANSFER: in acid/base chemistry this is used to denote the transfer of a hydrogen ion from an acid (H-donor) to an acceptor (Hacceptor). WEAK ACID or BASE VS. STRNG ACID or BASE: The strength of an acid or a base is a direct function of the degree of ionization of the acid or base (or, how well it does proton transfer reactions). A relative term. Acid-Base Reactions In the most general sense, an acid reacts with a base by transferring an ionizable hydrogen (an acidic H ) to it: The term conjugate is used exclusively to describe the materials on the product side of this reaction. Conjugates are easy to write: If the base is a hydroxide base, a neutralization reaction occurs: Slide 4-17 Slide 4-18

4 Acid Strength: Determining the Molarity of H+ Ions in an Aqueous Solution of a Strong Acid Strong Acids: HBr, HCl, HI, HN3, H2S4, HCl4 Strong Bases: NaH, KH, Ca(H)2, Sr(H)2, Ba(H)2, Group I and Group II oxides Sulfuric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4 M sulfuric acid? Slide 4-19 Slide 4-20 An acid-base titration. Writing Ionic Equations for Acid-Base Reactions Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions. (a) Strontium hydroxide(aq) + perchloric acid(aq) (b) Barium hydroxide(aq) + sulfuric acid(aq) Start of titration Excess of acid Slide 4-21 Finding the Concentration of Acid from an Acid-Base Titration You perform an acid-base titration to standardize an HCl solution by placing ml of HCl in a flask with a few drops of indicator solution. You put M NaH into the buret, and the initial reading is 0.55 ml. At the end point, the buret reading is ml. What is the concentration of the HCl solution? Slight excess of base Slide 4-22 SLUTIN: NaH (aq) + HCl (aq) NaCl (aq) + H2 (l) volume of base = ml 0.55 ml = ml ml soln x Strategy: Point of neutralization 1L 103 ml x mol NaH 1 L soln = 5.078x10-3 mol NaH Since 1 mol of HCl reacts with 1 mol NaH, the amount of HCl = 5.078x10-3 mol x10-3 mol HCl x 103 ml 1 L ml Slide 4-23 = M HCl Slide 4-24

5 xidation-reduction Reactions The redox process in compound formation. xidation: the LSS (via electron transfer) of one or more electrons: Reduction: the GAIN (via electron transfer) of one or more electrons: Slide 4-25 Slide 4-26 xidation Number (N) Determination 1. The N of an atom as an element is zero: Na(s), C(gr), H 2 (g), I 2 (s); 2. For a monatomic ion, its charge is its N; 3. F is always 1; 4. Group I elements are always +1; Group II elements are always +2; 5. xygen is usually assigned an N of 2; the only exception is for peroxides (species where you have bonds) where oxygen has an N of 1; 6. Hydrogen is +1 when combined with non-metals and 1 when combined with metals (hydrides); 7. The sum of the oxidation states of all atoms in a compound must equal the charge on the compound or ion of interest; 8. Non-integer oxidation states D exist (Fe 3 4 ) Slide 4-27 Slide 4-28 xidation Number (N) Determination A compound (ionic or covalent) is neutral and has a charge of zero. An ion WILL have a TTAL charge; The sum of all the oxidation states in a compound must equal the overall charge on the ion or molecule. Set up and solve for the unknown algebraically; For Fe 2 3 : [(# irons)! ( charge of iron) ] + (# oxygens)! ( charge of oxygen) ( 2x) ( ( )) For Cr : [ ] = charge of Fe 2 3 [ ] + [ 3 "2 ] = 0; solve to get x = +3 [(# chromiums)! ( charge of chromium) ] + (# oxygens)! ( charge of oxygen) ( 2x) ( ( )) 2 [ ] = charge of Cr 2 7 [ ] + [ 7 "2 ] = 2; solve to get x = +6 Balancing Redox Reactions: Half Reaction Method 1. Write separate equations for both the oxidation and reduction half reactions; 2. Balance under acidic conditions FIRST by doing the following for EACH half reaction separately: a. Balance all elements EXCEPT H and ; b. Balance using H 2 ; c. Balance H using H + ; d. Balance charge using electrons (e ) 3. Multiply each half reaction by an integer to ensure the number of electrons lost = number of electrons gained**; 4. Add half reactions and cancel species that occur on both the reactant AND product side; 5. Check to ensure balance! Slide 4-29 Slide 4-30

6 Balancing Redox Reactions: Basic Conditions 1. Balance under acidic conditions FIRST ; get overall balanced equation; 2. Add an equal number of H s as you have H + present to BTH sides of the equation; 3. When H + and H are present on the same side, make H 2 ; 4. Eliminate equal numbers of waters from both sides of the equation until only one side has waters left; 5. Check to ensure balance! Redox Balance: An Example Balance the following: + Mn 4 + Mn The half-reactions are: and Mn 4! "! Mn For the carbon species: H H e C balanced balanced H balanced charge balanced Slide 4-31 Slide 4-32 Redox Balance: An Example For the manganese species: Mn 4! "! Mn Mn 4 4 H + + Mn 4 3 e + 4 H + + Mn 4! "! Mn +! "! Mn +! "! Mn + Mn balanced balanced H balanced charge balanced Multiply the Mn equation by 2 to get correct number of electrons: H e 6 e + 8 H Mn 4! "! 2 Mn + 4 H 2 Slide 4-33 Redox Balance: An Example Combine reactions and eliminate common species: e H Mn 4 Final balanced equation: + 2 H Mn H e! "! + 2 Mn Mn + in base : + 2 H H + 2 Mn 4 overall : + 2 Mn 4! "! + 2 Mn H + 2 Mn + 2 H Slide 4-34 A summary of terminology for oxidation-reduction (redox) reactions. Highest and lowest oxidation numbers of reactive main-group elements. Slide 4-35 Slide 4-36

7 Recognizing xidizing and Reducing Agents Figure 4.12 An active metal displacing hydrogen from water. Identify the oxidizing agent and reducing agent in each of the following: (a) 2Al(s) + 3H2S4(aq) (b) Pb(s) + C(g) (c) 2H2(g) + 2(g) Al2(S4)3(aq) + 3H2(g) Pb(s) + C2(g) 2H2(g) Strategy: Assign an.n. for each atom and see which atom gained and which atom lost electrons in going from reactants to products. An increase in.n. means the species was oxidized (and is the reducing agent) and a decrease in.n. means the species was reduced (is the oxidizing agent) (a) 2Al(s) + 3H2S4(aq) Al2(S4)3(aq) + 3H2(g) The.N. of Al increases; Al is oxidized; it is the reducing agent. The.N. of H decreases; H is reduced; H2S4 is the oxidizing agent. Slide 4-37 Slide 4-38 Displacing one metal by another. The activity series of the metals. Slide 4-39 End of Chapter 4 Slide 4-41 Slide 4-40